# Thread: [SOLVED] Equation to plane through 3 pts

1. ## [SOLVED] Equation to plane through 3 pts

I have points p, q, r,

p = (-4,-1,-1)
q = (-2,0,1)
r = (-1,-2,-3)

Okay so I need to find the equation to the plane using these three points.

Step 1. I find vector pq and vector qr

vector pq = (2,1,2)
vector qr = (1,-2,-4)

Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)

Step 3. I sub this into any point using (I forget what it's called form) and I get

0(x+4) + 6(y+1) + -5(z+1) = 0
6y - 5z +1 = 0

I think that's right, what do you think?

2. Originally Posted by thekrown
I have points p, q, r,

p = (-4,-1,-1)
q = (-2,0,1)
r = (-1,-2,-3)

Okay so I need to find the equation to the plane using these three points.

Step 1. I find vector pq and vector qr

vector pq = (2,1,2)
vector qr = (1,-2,-4)

Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)

Step 3. I sub this into any point using (I forget what it's called form) and I get

0(x+4) + 6(y+1) + -5(z+1) = 0
6y - 5z +1 = 0

I think that's right, what do you think?
Neither q nor r are on that plane...

Tonio

3. What am I doing wrong? Are my steps correct or are my calculations wrong?

4. Originally Posted by thekrown
What am I doing wrong? Are my steps correct or are my calculations wrong?
$\overrightarrow {PQ} = \left\langle {2,1,2} \right\rangle \,\& \,\overrightarrow {PR} = \left\langle {3, - 1, - 2} \right\rangle$
$\overrightarrow {PQ} \times \,\overrightarrow {PR} = \left\langle {0,2, - 1} \right\rangle$

5. Originally Posted by thekrown
I have points p, q, r,

p = (-4,-1,-1)
q = (-2,0,1)
r = (-1,-2,-3)

Okay so I need to find the equation to the plane using these three points.

Step 1. I find vector pq and vector qr

vector pq = (2,1,2)
vector qr = (1,-2,-4)

Step 2. Find the cross product pq x qr since they are parallel to the plane. I got (0,6,-5)
pq x qr= (0, 10, -5)

Step 3. I sub this into any point using (I forget what it's called form) and I get

0(x+4) + 6(y+1) + -5(z+1) = 0
6y - 5z +1 = 0

I think that's right, what do you think?
You should at least have put p, q, and r into that equation to see for yourself!